YES We show the termination of the TRS R: app(app(|:|(),x),x) -> e() app(app(|:|(),e()),x) -> x app(app(|:|(),x),app(app(.(),x),y)) -> y app(app(|:|(),app(app(/(),x),y)),x) -> y app(app(/(),x),x) -> e() app(app(/(),x),e()) -> x app(app(/(),app(app(.(),y),x)),x) -> y app(app(/(),x),app(app(|:|(),y),x)) -> y app(app(.(),e()),x) -> x app(app(.(),x),e()) -> x app(app(.(),x),app(app(|:|(),x),y)) -> y app(app(.(),app(app(/(),y),x)),x) -> y app(app(map(),f),nil()) -> nil() app(app(map(),f),app(app(cons(),x),xs)) -> app(app(cons(),app(f,x)),app(app(map(),f),xs)) app(app(filter(),f),nil()) -> nil() app(app(filter(),f),app(app(cons(),x),xs)) -> app(app(app(app(filter2(),app(f,x)),f),x),xs) app(app(app(app(filter2(),true()),f),x),xs) -> app(app(cons(),x),app(app(filter(),f),xs)) app(app(app(app(filter2(),false()),f),x),xs) -> app(app(filter(),f),xs) Instead we prove termination of the uncurried system: app(._0(),y) -> ._1(y) app(._1(x1),y) -> ._2(x1,y) app(/_0(),y) -> /_1(y) app(/_1(x1),y) -> /_2(x1,y) app(cons_0(),y) -> cons_1(y) app(cons_1(x1),y) -> cons_2(x1,y) app(filter_0(),y) -> filter_1(y) app(filter_1(x1),y) -> filter_2(x1,y) app(filter2_0(),y) -> filter2_1(y) app(filter2_1(x1),y) -> filter2_2(x1,y) app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) app(map_0(),y) -> map_1(y) app(map_1(x1),y) -> map_2(x1,y) app(|:|_0(),y) -> |:|_1(y) app(|:|_1(x1),y) -> |:|_2(x1,y) |:|_2(x,x) -> e_0() |:|_2(e_0(),x) -> x |:|_2(x,._2(x,y)) -> y |:|_2(/_2(x,y),x) -> y /_2(x,x) -> e_0() /_2(x,e_0()) -> x /_2(._2(y,x),x) -> y /_2(x,|:|_2(y,x)) -> y ._2(e_0(),x) -> x ._2(x,e_0()) -> x ._2(x,|:|_2(x,y)) -> y ._2(/_2(y,x),x) -> y map_2(f,nil_0()) -> nil_0() map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) filter_2(f,nil_0()) -> nil_0() filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(._1(x1),y) -> ._2#(x1,y) p2: app#(/_1(x1),y) -> /_2#(x1,y) p3: app#(filter_1(x1),y) -> filter_2#(x1,y) p4: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p5: app#(map_1(x1),y) -> map_2#(x1,y) p6: app#(|:|_1(x1),y) -> |:|_2#(x1,y) p7: map_2#(f,cons_2(x,xs)) -> app#(f,x) p8: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p9: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p10: filter_2#(f,cons_2(x,xs)) -> app#(f,x) p11: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) p12: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs) and R consists of: r1: app(._0(),y) -> ._1(y) r2: app(._1(x1),y) -> ._2(x1,y) r3: app(/_0(),y) -> /_1(y) r4: app(/_1(x1),y) -> /_2(x1,y) r5: app(cons_0(),y) -> cons_1(y) r6: app(cons_1(x1),y) -> cons_2(x1,y) r7: app(filter_0(),y) -> filter_1(y) r8: app(filter_1(x1),y) -> filter_2(x1,y) r9: app(filter2_0(),y) -> filter2_1(y) r10: app(filter2_1(x1),y) -> filter2_2(x1,y) r11: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r12: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r13: app(map_0(),y) -> map_1(y) r14: app(map_1(x1),y) -> map_2(x1,y) r15: app(|:|_0(),y) -> |:|_1(y) r16: app(|:|_1(x1),y) -> |:|_2(x1,y) r17: |:|_2(x,x) -> e_0() r18: |:|_2(e_0(),x) -> x r19: |:|_2(x,._2(x,y)) -> y r20: |:|_2(/_2(x,y),x) -> y r21: /_2(x,x) -> e_0() r22: /_2(x,e_0()) -> x r23: /_2(._2(y,x),x) -> y r24: /_2(x,|:|_2(y,x)) -> y r25: ._2(e_0(),x) -> x r26: ._2(x,e_0()) -> x r27: ._2(x,|:|_2(x,y)) -> y r28: ._2(/_2(y,x),x) -> y r29: map_2(f,nil_0()) -> nil_0() r30: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r31: filter_2(f,nil_0()) -> nil_0() r32: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r33: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r34: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The estimated dependency graph contains the following SCCs: {p3, p4, p5, p7, p8, p9, p10, p11, p12} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(filter_1(x1),y) -> filter_2#(x1,y) p2: filter_2#(f,cons_2(x,xs)) -> app#(f,x) p3: app#(map_1(x1),y) -> map_2#(x1,y) p4: map_2#(f,cons_2(x,xs)) -> map_2#(f,xs) p5: map_2#(f,cons_2(x,xs)) -> app#(f,x) p6: app#(filter2_3(x1,x2,x3),y) -> filter2_4#(x1,x2,x3,y) p7: filter2_4#(false_0(),f,x,xs) -> filter_2#(f,xs) p8: filter_2#(f,cons_2(x,xs)) -> filter2_4#(app(f,x),f,x,xs) p9: filter2_4#(true_0(),f,x,xs) -> filter_2#(f,xs) and R consists of: r1: app(._0(),y) -> ._1(y) r2: app(._1(x1),y) -> ._2(x1,y) r3: app(/_0(),y) -> /_1(y) r4: app(/_1(x1),y) -> /_2(x1,y) r5: app(cons_0(),y) -> cons_1(y) r6: app(cons_1(x1),y) -> cons_2(x1,y) r7: app(filter_0(),y) -> filter_1(y) r8: app(filter_1(x1),y) -> filter_2(x1,y) r9: app(filter2_0(),y) -> filter2_1(y) r10: app(filter2_1(x1),y) -> filter2_2(x1,y) r11: app(filter2_2(x1,x2),y) -> filter2_3(x1,x2,y) r12: app(filter2_3(x1,x2,x3),y) -> filter2_4(x1,x2,x3,y) r13: app(map_0(),y) -> map_1(y) r14: app(map_1(x1),y) -> map_2(x1,y) r15: app(|:|_0(),y) -> |:|_1(y) r16: app(|:|_1(x1),y) -> |:|_2(x1,y) r17: |:|_2(x,x) -> e_0() r18: |:|_2(e_0(),x) -> x r19: |:|_2(x,._2(x,y)) -> y r20: |:|_2(/_2(x,y),x) -> y r21: /_2(x,x) -> e_0() r22: /_2(x,e_0()) -> x r23: /_2(._2(y,x),x) -> y r24: /_2(x,|:|_2(y,x)) -> y r25: ._2(e_0(),x) -> x r26: ._2(x,e_0()) -> x r27: ._2(x,|:|_2(x,y)) -> y r28: ._2(/_2(y,x),x) -> y r29: map_2(f,nil_0()) -> nil_0() r30: map_2(f,cons_2(x,xs)) -> cons_2(app(f,x),map_2(f,xs)) r31: filter_2(f,nil_0()) -> nil_0() r32: filter_2(f,cons_2(x,xs)) -> filter2_4(app(f,x),f,x,xs) r33: filter2_4(true_0(),f,x,xs) -> cons_2(x,filter_2(f,xs)) r34: filter2_4(false_0(),f,x,xs) -> filter_2(f,xs) The set of usable rules consists of r1, r3, r7, r9, r13, r17, r18, r19, r21, r22, r25, r27 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: app#_A(x1,x2) = 0 filter_1_A(x1) = x1 + 1 filter_2#_A(x1,x2) = 0 cons_2_A(x1,x2) = 1 map_1_A(x1) = x1 + 1 map_2#_A(x1,x2) = 0 filter2_3_A(x1,x2,x3) = x1 + x3 + 1 filter2_4#_A(x1,x2,x3,x4) = 0 false_0_A() = 1 app_A(x1,x2) = x1 + x2 + 1 true_0_A() = 1 ._0_A() = 1 ._1_A(x1) = x1 + 1 /_0_A() = 1 /_1_A(x1) = x1 + 1 filter_0_A() = 1 filter2_0_A() = 1 filter2_1_A(x1) = x1 + 1 map_0_A() = 1 |:|_2_A(x1,x2) = x1 + x2 + 1 e_0_A() = 1 ._2_A(x1,x2) = x1 + x2 /_2_A(x1,x2) = x1 + x2 + 1 app#!_A(x1,x2) = 8 filter_1!_A(x1) = 1 filter_2#!_A(x1,x2) = 8 cons_2!_A(x1,x2) = x1 + 9 map_1!_A(x1) = 1 map_2#!_A(x1,x2) = 8 filter2_3!_A(x1,x2,x3) = 4 filter2_4#!_A(x1,x2,x3,x4) = 8 false_0!_A() = 1 app!_A(x1,x2) = x2 + 7 true_0!_A() = 0 ._0!_A() = 0 ._1!_A(x1) = 1 /_0!_A() = 0 /_1!_A(x1) = 1 filter_0!_A() = 0 filter2_0!_A() = 7 filter2_1!_A(x1) = 6 map_0!_A() = 0 |:|_2!_A(x1,x2) = 9 e_0!_A() = 0 ._2!_A(x1,x2) = 2 /_2!_A(x1,x2) = 2 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: app#_A(x1,x2) = max{1, x1 - 17, x2 - 21} filter_1_A(x1) = x1 + 15 filter_2#_A(x1,x2) = max{1, x1 - 2, x2 - 36} cons_2_A(x1,x2) = max{x1 + 15, x2 + 35} map_1_A(x1) = max{16, x1 + 15} map_2#_A(x1,x2) = max{1, x1 - 17, x2 - 36} filter2_3_A(x1,x2,x3) = max{16, x1 + 15, x2 + 15, x3 + 15} filter2_4#_A(x1,x2,x3,x4) = max{1, x2 - 2, x3 - 21, x4 - 21} false_0_A = 2 app_A(x1,x2) = max{x1 - 17, x2 + 38} true_0_A = 2 ._0_A = 15 ._1_A(x1) = x1 + 15 /_0_A = 15 /_1_A(x1) = x1 + 15 filter_0_A = 15 filter2_0_A = 15 filter2_1_A(x1) = x1 + 15 map_0_A = 16 |:|_2_A(x1,x2) = max{3, x1 - 4, x2 + 1} e_0_A = 2 ._2_A(x1,x2) = max{38, x2} /_2_A(x1,x2) = max{3, x1 + 1, x2 - 1} app#!_A(x1,x2) = x2 + 7 filter_1!_A(x1) = x1 + 1 filter_2#!_A(x1,x2) = max{4, x2 - 7} cons_2!_A(x1,x2) = max{13, x1 - 39, x2 - 5} map_1!_A(x1) = x1 + 10 map_2#!_A(x1,x2) = max{4, x2 - 7} filter2_3!_A(x1,x2,x3) = 11 filter2_4#!_A(x1,x2,x3,x4) = max{6, x4 - 6} false_0!_A = 5 app!_A(x1,x2) = max{9, x1 - 15} true_0!_A = 3 ._0!_A = 9 ._1!_A(x1) = max{10, x1 - 2} /_0!_A = 1 /_1!_A(x1) = max{10, x1 - 2} filter_0!_A = 2 filter2_0!_A = 2 filter2_1!_A(x1) = 1 map_0!_A = 11 |:|_2!_A(x1,x2) = max{10, x1 - 1, x2 - 1} e_0!_A = 1 ._2!_A(x1,x2) = max{11, x1 - 1, x2 + 10} /_2!_A(x1,x2) = max{8, x1 - 1} partial status: pi(app#) = [] pi(filter_1) = [1] pi(filter_2#) = [] pi(cons_2) = [] pi(map_1) = [] pi(map_2#) = [] pi(filter2_3) = [] pi(filter2_4#) = [] pi(false_0) = [] pi(app) = [1, 2] pi(true_0) = [] pi(._0) = [] pi(._1) = [1] pi(/_0) = [] pi(/_1) = [1] pi(filter_0) = [] pi(filter2_0) = [] pi(filter2_1) = [] pi(map_0) = [] pi(|:|_2) = [1, 2] pi(e_0) = [] pi(._2) = [1, 2] pi(/_2) = [1, 2] The following rules are removed: p1, p2, p3, p4, p5, p6, p7, p8, p9 Then no dependency pair remains.